Question

Compare the time periods of two pendulums of length
1 m and 9 m
solve it .​

Answer 1

Given:-

• Length of first pendulum = $\sf l_1 = 1\ m$
• Length of second pendulum = $\sf l_2 = 9\ m$

To compare:-

• Time period of first pendulum $\sf (T_1)$ and time period of second pendulum $\sf (T_2).$

Formula to be used:-

• $\sf T = 2\pi \sqrt{\dfrac{l}{g}}$ where T is the time period of the pendulum, l is the length of the pendulum (measured from point where the pendulum is connected to the upper support till the centre of the Bob) and g is the acceleration due to gravity.

Answer:-

Finding time period of first pendulum:-

$\sf T_1 = 2\pi \sqrt{\dfrac{l_1}{g}}$

Putting the value of $\sf l_1 = 1\ m,$

$\longrightarrow \sf T_1 = 2\pi \sqrt{\dfrac{1}{g}}\quad \quad \dots (1)$

Finding time period of second pendulum:-

$\sf T_2 = 2\pi \sqrt{\dfrac{l_2}{g}}$

Putting the value of $\sf l_2 = 9\ m,$

$\longrightarrow \sf T_2 = 2\pi \sqrt{\dfrac{9}{g}}\quad \quad \dots (2)$

Finding the relation between the time periods:-

From (1) and (2),

$\sf \dfrac{T_1}{T_2} = \dfrac{ \bigg( 2\pi \sqrt{\dfrac{1}{g}} \bigg) }{ \bigg( 2\pi \sqrt{\dfrac{9}{g}}\bigg) }$

$\sf \longrightarrow \dfrac{T_1}{T_2} = \dfrac{ \bigg( \cancel{2\pi} \sqrt{\dfrac{1}{g}} \bigg) }{ \bigg( \cancel{2\pi} \sqrt{\dfrac{9}{g}}\bigg) }$

$\sf \longrightarrow \dfrac{T_1}{T_2} = \dfrac{\bigg( \dfrac{\sqrt{1}}{\cancel{\sqrt{g}}} \bigg)}{\bigg( \dfrac{\sqrt{9}}{\cancel{\sqrt{g}}} \bigg) }$

$\sf \longrightarrow \dfrac{T_1}{T_2} = \dfrac{\sqrt{1}}{\sqrt{9}}$

$\sf \longrightarrow \dfrac{T_1}{T_2} = \dfrac{1}{3}$

$\longrightarrow \underline{\underline{\sf{\green{3T_1 = T_2}}}}$